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GarryVolchara [31]
3 years ago
5

An object speeding up, an object slowing down, and an object changing direction are all examples of acceleration.

Physics
1 answer:
emmainna [20.7K]3 years ago
6 0
<h2>Answer: True</h2>

When an object is speeding up, the acceleration is in the same direction as the velocity. Thus, this object has a positive acceleration. In Example B, the object is moving in the negative direction (i.e., has a negative velocity) and is slowing down.

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Write a program that calculates the average of n integers. the program should prompt the user to enter the value for n and then
ikadub [295]

#include<studio.h>

int main( )

{

int n;

int a,b,c,d,x,y;

int avarage;

printf("enter value of n:\n");

scanf("%d",&n);

printf("enter value of a:\n,b:\n,c:\n,d:\n,x:\n,y:\n);

scan f("%d\%d\n%d\n%d\n%d\n%d\n",&a,b,c,d,x,y);

sum=(a+b+c+d+x+y);

avarage=(sum/n);

print f("%d",avarage);

if

{

n=positive interger

}

else

{

printf ("n must be positive");

}

return 0;

}

8 0
3 years ago
atmospheric pressure is measured in all of the following units except ……… ? A. torr B. pascal C. bar D. newton?
AlexFokin [52]

Answer:

C. Bar

Explanation:

6 0
2 years ago
An object has a velocity of 8 m/s and a kinetic energy of 480 J. What is the mass of the object? (Formula: )
andrezito [222]

Answer:

1/2mv^2 a(7.5 b(15 kg c(60 kg d(120 kg. nevermind i found the answer its (15 kg) because to solve for m its m= K2/v squared.

Explanation:

6 0
3 years ago
The diagrams show objects' gravitational pull toward each
zaharov [31]
I don’t know sorry then ya soo
4 0
3 years ago
(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

3 0
3 years ago
Read 2 more answers
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